A263133 -id:A263133 - cp's OEIS frontend

A007228 a(n) = 3binomial(4n,n)/(n+1).

3, 6, 28, 165, 1092, 7752, 57684, 444015, 3506100, 28242984, 231180144, 1917334783, 16077354108, 136074334200, 1160946392760, 9973891723635, 86210635955220, 749191930237608, 6541908910355280, 57369142749576660, 505045163173167760, 4461713825057817120

Offset: 0

Simon Plouffe

For n >= 1, a(n) is the number of distinct perforation patterns for deriving (v,b) = (n+1,n) punctured convolutional codes from (4,1). [Edited by Petros Hadjicostas, Jul 27 2020]
Apparently Bégin's (1992) paper was presented at a poster session at the conference and was never published.
a(n) is the total number of down steps between the first and second up steps in all 3-Dyck paths of length 4*(n+1). A 3-Dyck path is a nonnegative lattice path with steps (1,3), (1,-1) that starts and ends at y = 0. - Sarah Selkirk, May 07 2020
From Petros Hadjicostas, Jul 27 2020: (Start)
"A punctured convolutional code is a high-rate code obtained by the periodic elimination (i.e., puncturing) of specific code symbols from the output of a low-rate encoder. The resulting high-rate code depends on both the low-rate code, called the original code, and the number and specific positions of the punctured symbols." (The quote is from Haccoun and Bégin (1989).)
A high-rate code (v,b) (written as R = b/v) can be constructed from a low-rate code (v0,1) (written as R = 1/v0) by deleting from every v0*b code symbols a number of v0*b - v symbols (so that the resulting rate is R = b/v).
Even though my formulas below do not appear in the two published papers in the IEEE Transactions on Communications, from the theory in those two papers, it makes sense to replace "k|b" with "k|v0*b" (and "k|gcd(v,b)" with "k|gcd(v,v0*b)"). Pab Ter, however, uses "k|b" in the Maple programs in the related sequences A007223, A007224, A007225, A007227, and A007229. (End)
Conjecture: for n >= 1, a(n) is odd iff n = 4*A263133(k) + 3 for some k. - Peter Bala, Mar 13 2023

			From _Petros Hadjicostas_, Jul 29 2020: (Start)
We give some examples to illustrate the comment by _Sarah Selkirk_ about the total number of downs between the 1st and 2nd ups in a 2-Dyck path of length 4*(n+1). We denote by (+3) an up movement by a vector of (1,3) and by (-1) a down movement by a vector of (1,-1). We use powers to denote repetition of the same movement.
(i) For n = 0, we have the following 2-Dyck path of length 4 that contributes to a(0) = 3: (+3)(-1)^3 (no 2nd up here) with a total of 3 downs after the 1st up.
(ii) For n = 1, we have the following 2-Dyck paths of length 8 that contribute to a(1) = 6: (+3)(-1)(+3)(-1)^5, (+3)(-1)^2(+3)(-1)^4, and (+3)(-1)^3(+3)(-1)^3 with a contribution of 1 + 2 + 3 = 6 downs between the 1st and 2nd ups.
(iii) For n = 2, we have the following 2-Dyck paths of length 12 that contribute to a(2) = 28: (+3)(-1)(+3)(-1)^i(+3)(-1)^(8-i) for i = 0..5, (+3)(-1)^2(+3)(-1)^i(+3)^(7-i) for i = 0..4, and (+3)(-1)^3(+3)(-1)^i(+3)(-1)^(6-i) for i = 0..3 with a contribution of 1 x 6 + 2 x 5 + 3 x 4 = 28 downs between the 1st and 2nd ups. (End)

Guy Bégin, On the enumeration of perforation patterns for punctured convolutional codes, Séries Formelles et Combinatoire Algébrique, 4th colloquium, 15-19 Juin 1992, Montréal, Université du Québec à Montréal, pp. 1-10.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Andrew Howroyd, Table of n, a(n) for n = 0..500
A. Asinowski, B. Hackl, and S. Selkirk, Down step statistics in generalized Dyck paths, arXiv:2007.15562 [math.CO], 2020.
Guy Bégin and David Haccoun, High rate punctured convolutions codes: Structure properties and construction techniques, IEEE Transactions on Communications 37(12) (1989), 1381-1385.
Fabio Deelan Cunden, Marilena Ligabò, and Tommaso Monni, Random matrices associated to Young diagrams, arXiv:2301.13555 [math.PR], 2023. See p. 7.
N. S. S. Gu, H. Prodinger, and S. Wagner, Bijections for a class of labeled plane trees, Eur. J. Combinat. 31 (2010), 720-732, Theorem 2 at k=3.
David Haccoun and Guy Bégin, High rate punctured convolutional codes for Viterbi and sequential coding, IEEE Transactions on Communications, 37(11) (1989), 1113-1125; see Section II.

Cf. A002293, A007223, A007224, A007225, A007226, A007227, A007229, A124724, A263133, A006632.

[3*Binomial(4*n,n)/(n+1) : n in [0..25]]; // Wesley Ivan Hurt, Jul 27 2020

Table[3/(n+1) Binomial[4n,n],{n,0,30}] (* Harvey P. Dale, Nov 14 2013 *)

a(n)={3*binomial(4*n,n)/(n+1)} \\ Andrew Howroyd, May 08 2020

a(n) = C(4*n,n)/(3*n+1) + 2*C(4*n+1,n)/(3*n+2) + 3*C(4*n+2,n)/(3*n+3). - Paul Barry, Nov 05 2006
G.f.: g + g^2 + g^3 where g = 1 + x*g^4 is the g.f. of A002293. - Mark van Hoeij, Nov 11 2011
3*(3*n-1)*(3*n-2)*(n+1)*a(n) - 8*(4*n-3)*(2*n-1)*(4*n-1)*a(n-1) = 0. - R. J. Mathar, Nov 24 2012
From Petros Hadjicostas, Jul 27 2020: (Start)
The number of perforation patterns to derive high-rate convolutional code (v,b) (written as R = b/v) from a given low-rate convolutional code (v0, 1) (written as R = 1/v0) is (1/b)*Sum_{k|gcd(v,b)} phi(k)*binomial(v0*b/k, v/k).
According to Pab Ter's Maple code in the related sequences (see above), this is the coefficient of z^v in the polynomial (1/b)*Sum_{k|b} phi(k)*(1 + z^k)^(v0*b/k).
Here (v,b) = (n+1,n) and (v0,1) = (4,1), so for n >= 1,
a(n) = (1/n)*Sum_{k|gcd(n+1,n)} phi(k)*binomial(4*n/k, (n+1)/k).
This simplifies to
a(n) = (1/n)*binomial(4*n, n+1) for n >= 1. (End)

Edited by N. J. A. Sloane, Feb 07 2004 following a suggestion of Ralf Stephan

Reedited by N. J. A. Sloane, May 31 2008 following a suggestion of R. J. Mathar

A263132 Positive values of m such that binomial(4*m - 1, m) is odd.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 11, 12, 16, 22, 24, 32, 43, 44, 48, 64, 86, 88, 96, 128, 171, 172, 176, 192, 256, 342, 344, 352, 384, 512, 683, 684, 688, 704, 768, 1024, 1366, 1368, 1376, 1408, 1536, 2048, 2731, 2732, 2736, 2752, 2816, 3072, 4096, 5462, 5464, 5472, 5504

Offset: 1

Peter Bala, Oct 10 2015

This sequence, when viewed as a set, equals the set of numbers of the form 4^n * ceiling(2^k/3) for n >= 0, k >= 1, i.e., the product subset in Z of A000302 and A005578 regarded as sets. See the example below.
Equivalently, this sequence, when viewed as a set, equals the set of numbers of the form 2^n * (2^(2*k + 1) + 1)/3 for n,k >= 0, i.e., the product subset in Z of A000079 and A007583 regarded as sets. See the example below.
2*a(n) gives the values of m such that binomial(4*m - 2,m) is odd. 4*a(n) gives the values of m such that binomial(4*m - 3,m) is odd (other than m = 1) and also the values of m such that binomial(4*m - 4,m) is odd.

			1) Notice how this sequence can be read from Table 1 below by moving through the table in a sequence of 'knight moves' (1 down and 2 to the left) starting from the first row. For example, starting at 11 on the top row we move in a series of knights moves 11 -> 12 -> 16, then return to the top row at 22 and move 22 -> 24 -> 32, return to the top row at 43 and move 43 -> 44 -> 48 -> 64, then return to top row at 86 and so on.
........................................................
.   Table 1: 4^n * ceiling(2^k/3) for n >= 0, k >= 1   .
........................................................
n\k|   1    2    3    4     5     6    7    8     9
---+----------------------------------------------------
0  |   1    2    3    6    11    22   43   86   171 ...
1  |   4    8   12   24    44    88  172  ...
2  |  16   32   48   96   176    ...
3  |  64  128  192  ...
4  | 256  ...
...
2) Notice how this sequence can be read from Table 2 below in a sequence of 'knight moves' (2 down and 1 to the left) starting from the first two rows. For example, starting at 43 in the first row we jump 43 -> 44 -> 48 -> 64, then return to the second row at 86 and jump 86 -> 88 -> 96 -> 128, followed by 171 -> 172 -> 176 -> 192 -> 256, and so on.
....................................................
.   Table 2: 2^n * (2^(2*k + 1) + 1)/3, n,k >= 0   .
....................................................
n\k|   0    1     2     3      4      5
---+----------------------------------------------
0  |   1    3    11    43    171    683  ...
1  |   2    6    22    86    342   1366  ...
2  |   4   12    44   172    684   2732  ...
3  |   8   24    88   344   1368   5464  ...
4  |  16   48   176   688   2736  10928  ...
5  |  32   96   352  1376   5472  21856  ...
6  |  64  192   704  2752  10944  43712  ...
7  | 128  384  1408  5504  21888  87424  ...
8  | 256 ...

Cf. A000079, A000267, A000302, A005578, A007583, A048716, A384688.

Other odd binomials: A002450 (4*m+1,m), A020988 (4*m+2,m), A263133 (4*m+3,m), A080674 (4*m+4,m), A118113 (3*m-2,m), A003714 (3*m,m).

[n: n in [1..6000] | Binomial(4*n-1, n) mod 2 eq 1]; // Vincenzo Librandi, Oct 12 2015

for n from 1 to 5000 do if mod(binomial(4*n-1, n), 2) = 1 then print(n) end if end do;

Select[Range[6000],OddQ[Binomial[4#-1,#]]&] (* Harvey P. Dale, Dec 26 2015 *)

for(n=1, 1e4, if (binomial(4*n-1, n) % 2 == 1, print1(n", "))) \\ Altug Alkan, Oct 11 2015

a(n) = my(r,s=sqrtint(4*n-3,&r)); (1<Kevin Ryde, Jun 14 2025

Python

A263132_list = [m for m in range(1,10**6) if not ~(4*m-1) & m] # Chai Wah Wu, Feb 07 2016

Formula

a(n) = A263133(n) + 1.

m is a term if and only if m AND NOT (4*m-1) = 0 where AND and NOT are bitwise operators. - Chai Wah Wu, Feb 07 2016

a(n) = (2^A000267(n-1) + 2^A384688(n-1)) / 3. - Kevin Ryde, Jun 14 2025

Extensions

More terms from Vincenzo Librandi, Oct 12 2015

cp's OEIS Frontend

A007228 a(n) = 3*binomial(4*n,n)/(n+1).

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A263132 Positive values of m such that binomial(4*m - 1, m) is odd.

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A369083 Expansion of g.f. A(x) satisfying A(x) = 1 + x*(5*A(x)^2 - A(-x)^2)/4.

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A007228 a(n) = 3binomial(4n,n)/(n+1).

A369083 Expansion of g.f. A(x) satisfying A(x) = 1 + x(5A(x)^2 - A(-x)^2)/4.